Let us have a surface given by the equation kind
Let us introduce the following definition.
Definition 1. A straight line is called tangent to the surface at some point if it is
tangent to any curve lying on the surface and passing through the point.
Since an infinite number of different curves lying on the surface pass through the point P, then, generally speaking, there will be an infinite number of tangents to the surface passing through this point.
Let us introduce the concept of singular and ordinary points of a surface
If at a point all three derivatives are equal to zero or at least one of these derivatives does not exist, then the point M is called a singular point of the surface. If at a point all three derivatives exist and are continuous, and at least one of them is different from zero, then the point M is called an ordinary point of the surface.
Now we can formulate the following theorem.
Theorem. All tangent lines to a given surface (1) at its ordinary point P lie in the same plane.
Proof. Let us consider a certain line L on the surface (Fig. 206) passing through a given point P of the surface. Let the curve under consideration be given by parametric equations
The tangent to the curve will be the tangent to the surface. The equations of this tangent have the form
If expressions (2) are substituted into equation (1), then this equation will turn into an identity with respect to t, since curve (2) lies on surface (1). Differentiating it by we get
The projections of this vector depend on - the coordinates of point P; note that since point P is ordinary, these projections at point P do not simultaneously vanish and therefore
tangent to a curve passing through point P and lying on the surface. The projections of this vector are calculated based on equations (2) at the value of the parameter t corresponding to point P.
Let us calculate the scalar product of vectors N and which is equal to the sum of the products of projections of the same name:
Based on equality (3), the expression on the right side is equal to zero, therefore,
From the last equality it follows that the vector LG and the tangent vector to curve (2) at point P are perpendicular. The above reasoning is valid for any curve (2) passing through point P and lying on the surface. Consequently, each tangent to the surface at point P is perpendicular to the same vector N and therefore all these tangents lie in the same plane perpendicular to the vector LG. The theorem has been proven.
Definition 2. The plane in which all tangent lines to the lines on the surface passing through its given point P are located is called the tangent plane to the surface at point P (Fig. 207).
Note that at singular points of the surface there may not be a tangent plane. At such points, the tangent lines to the surface may not lie in the same plane. For example, the vertex of a conical surface is a singular point.
The tangents to the conical surface at this point do not lie in the same plane (they themselves form a conical surface).
Let us write the equation of the tangent plane to the surface (1) at an ordinary point. Since this plane is perpendicular to vector (4), therefore, its equation has the form
If the equation of the surface is given in the form or the equation of the tangent plane in this case takes the form
Comment. If we put in formula (6), then this formula will take the form
its right side is full differential functions Hence, . Thus, the total differential of a function of two variables at a point corresponding to the increments of the independent variables x and y is equal to the corresponding increment of the applicate of the tangent plane to the surface, which is the graph of this function.
Definition 3. A straight line drawn through a point on the surface (1) perpendicular to the tangent plane is called the normal to the surface (Fig. 207).
Let's write the normal equations. Since its direction coincides with the direction of vector N, its equations will have the form
1°1°. Equations of the tangent plane and normal for the case of explicit definition of the surface.
Let's consider one of the geometric applications of partial derivatives of a function of two variables. Let the function z = f (x ;y) differentiable at the point (x 0; y 0) some area DÎ R 2. Let's cut the surface S, representing the function z, planes x = x 0 And y = y 0(Fig. 11).
Plane X = x 0 intersects the surface S along some line z 0 (y ), the equation of which is obtained by substituting into the expression of the original function z ==f (x ;y) instead of X numbers x 0 . Dot M 0 (x 0 ;y 0,f (x 0 ;y 0)) belongs to the curve z 0 (y). Due to the differentiable function z at the point M 0 function z 0 (y) is also differentiable at the point y =y 0 . Therefore, at this point in the plane x = x 0 to the curve z 0 (y) a tangent can be drawn l 1.
Carrying out similar reasoning for the section at = y 0, let's build a tangent l 2 to the curve z 0 (x) at the point X = x 0 - Direct 1 1 And 1 2 define a plane called tangent plane to the surface S at the point M 0.
Let's create its equation. Since the plane passes through the point Mo(x 0 ;y 0 ;z 0), then its equation can be written as
A(x - xo) + B(y - yo) + C (z - zo) = 0,
which can be rewritten like this:
z -z 0 = A 1 (x – x 0) + B 1 (y – y 0) (1)
(dividing the equation by -C and denoting 
).
We'll find A 1 and B 1.
Tangent equations 1 1 And 1 2 look like
respectively.
Tangent l 1 lies in plane a , therefore, the coordinates of all points l 1 satisfy equation (1). This fact can be written in the form of a system
Resolving this system with respect to B 1, we obtain that. Carrying out similar reasoning for the tangent l 3, it is easy to establish that .
Substituting the values A 1 and B 1 into equation (1), we obtain the required tangent plane equation:
Line passing through a point M 0 and perpendicular to the tangent plane constructed at this point on the surface is called its normal.
Using the condition of perpendicularity of the line and the plane, it is easy to obtain the canonical normal equations:
Comment. Formulas for the tangent plane and normal to the surface are obtained for ordinary, i.e., non-special, points of the surface. Dot M 0 surface is called special, if at this point all partial derivatives are equal to zero or at least one of them does not exist. We do not consider such points.
Example. Write equations for the tangent plane and normal to the surface at its point M(2; -1; 1).
Solution. Let's find the partial derivatives of this function and their values at point M
From here, applying formulas (2) and (3), we will have: z-1=2(x-2)+2(y+1) or 2х+2у-z-1=0- tangent plane equation and 
- normal equations.
2°. Equations of the tangent plane and normal for the case of implicit definition of the surface.
If the surface S given by the equation F (x ; y;z)= 0, then equations (2) and (3), taking into account the fact that partial derivatives can be found as derivatives of an implicit function.
A surface is defined as a set of points whose coordinates satisfy a certain type of equation:
F (x , y , z) = 0 (1) (\displaystyle F(x,\,y,\,z)=0\qquad (1))If the function F (x , y , z) (\displaystyle F(x,\,y,\,z)) is continuous at some point and has continuous partial derivatives at it, at least one of which does not vanish, then in the neighborhood of this point the surface given by equation (1) will be the right surface.
In addition to the above implicit way of specifying, the surface can be defined obviously, if one of the variables, for example, z, can be expressed in terms of the others:
z = f (x , y) (1 ′) (\displaystyle z=f(x,y)\qquad (1"))More strictly simple surface is called the image of a homeomorphic mapping (that is, a one-to-one and mutually continuous mapping) of the interior of a unit square. This definition can be given an analytical expression.
Let a square be given on a plane with a rectangular coordinate system u and v, the coordinates of the internal points of which satisfy the inequalities 0< u < 1, 0 < v < 1. Гомеоморфный образ квадрата в пространстве с прямоугольной системой координат х, у, z задаётся при помощи формул х = x(u, v), у = y(u, v), z = z(u, v) (параметрическое задание поверхности). При этом от функций x(u, v), y(u, v) и z(u, v) требуется, чтобы они были непрерывными и чтобы для различных точек (u, v) и (u", v") были различными соответствующие точки (x, у, z) и (x", у", z").
Example simple surface is a hemisphere. The entire sphere is not simple surface. This necessitates further generalization of the concept of surface.
A subset of space, each point of which has a neighborhood that is simple surface, called the right surface .
Surface in differential geometry
Helicoid
Catenoid
The metric does not uniquely determine the shape of the surface. For example, the metrics of a helicoid and a catenoid, parameterized accordingly, coincide, that is, there is a correspondence between their regions that preserves all lengths (isometry). Properties that are preserved under isometric transformations are called internal geometry surfaces. The internal geometry does not depend on the position of the surface in space and does not change when it is bent without tension or compression (for example, when a cylinder is bent into a cone).
Metric coefficients E , F , G (\displaystyle E,\ F,\ G) determine not only the lengths of all curves, but also in general the results of all measurements inside the surface (angles, areas, curvature, etc.). Therefore, everything that depends only on the metric refers to internal geometry.
Normal and normal section
Normal vectors at surface points
One of the main characteristics of a surface is its normal- unit vector perpendicular to the tangent plane at a given point:
m = [ r u ′ , r v ′ ] | [ r u ′ , r v ′ ] | (\displaystyle \mathbf (m) =(\frac ([\mathbf (r"_(u)) ,\mathbf (r"_(v)) ])(|[\mathbf (r"_(u)) ,\mathbf (r"_(v)) ]|))).The sign of the normal depends on the choice of coordinates.
A section of a surface by a plane containing the surface normal at a given point forms a certain curve called normal section surfaces. The main normal for a normal section coincides with the normal to the surface (up to sign).
If the curve on the surface is not a normal section, then its main normal forms a certain angle with the normal of the surface θ (\displaystyle \theta ). Then the curvature k (\displaystyle k) curve related to curvature k n (\displaystyle k_(n)) normal section (with the same tangent) by Meunier's formula:
k n = ± k cos θ (\displaystyle k_(n)=\pm k\,\cos \,\theta )Coordinates of the normal unit vector for different ways surface assignments are given in the table:
| Normal coordinates at a surface point | |
|---|---|
| implicit assignment | (∂ F ∂ x ; ∂ F ∂ y ; ∂ F ∂ z) (∂ F ∂ x) 2 + (∂ F ∂ y) 2 + (∂ F ∂ z) 2 (\displaystyle (\frac (\left(( \frac (\partial F)(\partial x));\,(\frac (\partial F)(\partial y));\,(\frac (\partial F)(\partial z))\right) )(\sqrt (\left((\frac (\partial F)(\partial x))\right)^(2)+\left((\frac (\partial F)(\partial y))\right) ^(2)+\left((\frac (\partial F)(\partial z))\right)^(2))))) |
| explicit assignment | (− ∂ f ∂ x ; − ∂ f ∂ y ; 1) (∂ f ∂ x) 2 + (∂ f ∂ y) 2 + 1 (\displaystyle (\frac (\left(-(\frac (\partial f )(\partial x));\,-(\frac (\partial f)(\partial y));\,1\right))(\sqrt (\left((\frac (\partial f)(\ partial x))\right)^(2)+\left((\frac (\partial f)(\partial y))\right)^(2)+1)))) |
| parametric specification | (D (y, z) D (u, v) ; D (z, x) D (u, v) ; D (x, y) D (u, v)) (D (y, z) D (u , v)) 2 + (D (z , x) D (u , v)) 2 + (D (x , y) D (u , v)) 2 (\displaystyle (\frac (\left((\frac (D(y,z))(D(u,v)));\,(\frac (D(z,x))(D(u,v)));\,(\frac (D(x ,y))(D(u,v)))\right))(\sqrt (\left((\frac (D(y,z))(D(u,v)))\right)^(2 )+\left((\frac (D(z,x))(D(u,v)))\right)^(2)+\left((\frac (D(x,y))(D( u,v)))\right)^(2))))) |
Here D (y , z) D (u , v) = | y u ′ y v ′ z u ′ z v ′ | , D (z , x) D (u , v) = | z u ′ z v ′ x u ′ x v ′ | , D (x, y) D (u, v) = | x u ′ x v ′ y u ′ y v ′ | (\displaystyle (\frac (D(y,z))(D(u,v)))=(\begin(vmatrix)y"_(u)&y"_(v)\\z"_(u) &z"_(v)\end(vmatrix)),\quad (\frac (D(z,x))(D(u,v)))=(\begin(vmatrix)z"_(u)&z" _(v)\\x"_(u)&x"_(v)\end(vmatrix)),\quad (\frac (D(x,y))(D(u,v)))=(\ begin(vmatrix)x"_(u)&x"_(v)\\y"_(u)&y"_(v)\end(vmatrix))).
All derivatives are taken at the point (x 0 , y 0 , z 0) (\displaystyle (x_(0),y_(0),z_(0))).
Curvature
For different directions at a given point on the surface, different curvature of the normal section is obtained, which is called normal curvature; it is assigned a plus sign if the main normal of the curve goes in the same direction as the normal to the surface, or a minus sign if the directions of the normals are opposite.
Generally speaking, at every point on a surface there are two perpendicular directions e 1 (\displaystyle e_(1)) And e 2 (\displaystyle e_(2)), in which the normal curvature takes minimum and maximum values; these directions are called main. The exception is the case when the normal curvature in all directions is the same (for example, near a sphere or at the end of an ellipsoid of revolution), then all directions at a point are principal.
Surfaces with negative (left), zero (center) and positive (right) curvature.
Normal curvatures in the principal directions are called main curvatures; let's designate them κ 1 (\displaystyle \kappa _(1)) And κ 2 (\displaystyle \kappa _(2)). Size:
K = κ 1 κ 2 (\displaystyle K=\kappa _(1)\kappa _(2))The graph of a function of 2 variables z = f(x,y) is a surface projected onto the XOY plane into the domain of definition of the function D.
Consider the surface σ
, given by the equation z = f(x,y), where f(x,y) is a differentiable function, and let M 0 (x 0 ,y 0 ,z 0) be a fixed point on the surface σ, i.e. z 0 = f(x 0 ,y 0). Purpose. The online calculator is designed to find tangent plane and surface normal equations. The solution is drawn up in Word format. If you need to find the equation of a tangent to a curve (y = f(x)), then you need to use this service.
Rules for entering functions:
Rules for entering functions:
- All variables are expressed through x,y,z
Tangent plane to the surface σ
at her point M 0 is the plane in which the tangents to all curves drawn on the surface lie σ
through the point M 0 .
The equation of the tangent plane to the surface defined by the equation z = f(x,y) at the point M 0 (x 0 ,y 0 ,z 0) has the form:
z – z 0 = f’ x (x 0 ,y 0)(x – x 0) + f’ y (x 0 ,y 0)(y – y 0)
The vector is called the surface normal vector σ at point M 0. The normal vector is perpendicular to the tangent plane.
Normal to surface σ at the point M 0 is a straight line passing through this point and having the direction of the vector N.
The canonical equations of the normal to the surface defined by the equation z = f(x,y) at the point M 0 (x 0 ,y 0 ,z 0), where z 0 = f(x 0 ,y 0), have the form:
Example No. 1. The surface is given by the equation x 3 +5y. Find the equation of the tangent plane to the surface at the point M 0 (0;1).
Solution. Let us write the tangent equations in general view: z - z 0 = f" x (x 0 ,y 0 ,z 0)(x - x 0) + f" y (x 0 ,y 0 ,z 0)(y - y 0)
According to the conditions of the problem, x 0 = 0, y 0 = 1, then z 0 = 5
Let's find the partial derivatives of the function z = x^3+5*y:
f" x (x,y) = (x 3 +5 y)" x = 3 x 2
f" x (x,y) = (x 3 +5 y)" y = 5
At point M 0 (0,1) the values of partial derivatives are:
f" x (0;1) = 0
f" y (0;1) = 5
Using the formula, we obtain the equation of the tangent plane to the surface at point M 0: z - 5 = 0(x - 0) + 5(y - 1) or -5 y+z = 0
Example No. 2. The surface is defined implicitly y 2 -1/2*x 3 -8z. Find the equation of the tangent plane to the surface at the point M 0 (1;0;1).
Solution. Finding the partial derivatives of a function. Since the function is specified implicitly, we look for derivatives using the formula: 
For our function:
Then: 
At point M 0 (1,0,1) values of partial derivatives:
f" x (1;0;1) = -3 / 16
f" y (1;0;1) = 0
Using the formula, we obtain the equation of the tangent plane to the surface at point M 0: z - 1 = -3 / 16 (x - 1) + 0(y - 0) or 3 / 16 x+z- 19 / 16 = 0
Example. Surface σ
given by the equation z= y/x + xy – 5x 3. Find the equation of the tangent plane and normal to the surface σ
at the point M 0 (x 0 ,y 0 ,z 0), belonging to her, if x 0 = –1, y 0 = 2.
Let's find the partial derivatives of the function z= f(x,y) = y/x + xy – 5x 3:
f x ’( x,y) = (y/x + xy – 5x 3)’ x = – y/x 2 + y – 15x 2 ;
f y ’ ( x,y) = (y/x + xy – 5x 3)’ y = 1/x + x.
Dot M 0 (x 0 ,y 0 ,z 0) belongs to the surface σ
, so we can calculate z 0 , substituting the given x 0 = –1 and y 0 = 2 into the surface equation:
z= y/x + xy – 5x 3
z 0 = 2/(-1) + (–1) 2 – 5 (–1) 3 = 1.At the point M 0 (–1, 2, 1) partial derivative values:
f x ’( M 0) = –1/(-1) 2 + 2 – 15(–1) 2 = –15; f y ’( M 0) = 1/(-1) – 1 = –2.
Using formula (5) we obtain the equation of the tangent plane to the surface σ at the point M 0:
z – 1= –15(x + 1) – 2(y – 2) z – 1= –15x – 15 – 2y + 4 15x + 2y + z + 10 = 0.
Using formula (6) we obtain the canonical equations of the normal to the surface σ at the point M 0:
.
Answers: tangent plane equation: 15 x + 2y + z+ 10 = 0; normal equations:
.
Example No. 1. Given a function z=f(x,y) and two points A(x 0, y 0) and B(x 1, y 1). Required: 1) calculate the value z 1 of the function at point B; 2) calculate the approximate value z 1 of the function at point B based on the value z 0 of the function at point A, replacing the increment of the function when moving from point A to point B with a differential; 3) create an equation for the tangent plane to the surface z = f(x,y) at point C(x 0 ,y 0 ,z 0).
Solution.
Let us write the tangent equations in general form:
z - z 0 = f" x (x 0 ,y 0 ,z 0)(x - x 0) + f" y (x 0 ,y 0 ,z 0)(y - y 0)
According to the conditions of the problem, x 0 = 1, y 0 = 2, then z 0 = 25
Let's find the partial derivatives of the function z = f(x,y)x^2+3*x*y*+y^2:
f" x (x,y) = (x 2 +3 x y +y 2)" x = 2 x+3 y 3
f" x (x,y) = (x 2 +3 x y +y 2)" y = 9 x y 2
At point M 0 (1,2) the values of partial derivatives are:
f" x (1;2) = 26
f" y (1;2) = 36
Using the formula, we obtain the equation of the tangent plane to the surface at point M 0:
z - 25 = 26(x - 1) + 36(y - 2)
or
-26 x-36 y+z+73 = 0
Example No. 2. Write the equations of the tangent plane and normal to the elliptic paraboloid z = 2x 2 + y 2 at the point (1;-1;3).
Namely, about what you see in the title. Essentially, this is a “spatial analogue” tangent finding problems And normals to the graph of a function of one variable, and therefore no difficulties should arise.
Let's start with the basic questions: WHAT IS a tangent plane and WHAT IS a normal? Many people understand these concepts at the level of intuition. The simplest model that comes to mind is a ball on which lies a thin flat piece of cardboard. The cardboard is located as close as possible to the sphere and touches it at a single point. In addition, at the point of contact it is secured with a needle sticking straight up.
In theory, there is a rather ingenious definition of a tangent plane. Imagine a free surface and the point belonging to it. Obviously, a lot passes through the point spatial lines, which belong to this surface. Who has what associations? =) ...personally, I imagined an octopus. Let us assume that each such line has spatial tangent at point .
Definition 1: tangent plane to the surface at a point - this is plane, containing tangents to all curves that belong to a given surface and pass through the point.
Definition 2: normal to the surface at a point - this is straight, passing through a given point perpendicular to the tangent plane.
Simple and elegant. By the way, so that you don’t die of boredom from the simplicity of the material, a little later I will share with you one elegant secret that allows you to forget about cramming various definitions ONCE AND FOR ALL.
Let's get acquainted with the working formulas and solution algorithm using a specific example. In the vast majority of problems, it is necessary to construct both the tangent plane equation and the normal equation:
Example 1
Solution:if the surface is given by the equation (i.e. implicitly), then the equation of the tangent plane to a given surface at a point can be found using the following formula:
Special attention I draw attention to unusual partial derivatives - their should not be confused With partial derivatives of an implicitly specified function (although the surface is specified implicitly). When finding these derivatives, one must be guided by rules for differentiating a function of three variables, that is, when differentiating with respect to any variable, the other two letters are considered constants:
Without leaving the cash register, we find the partial derivative at the point:
Likewise: 
This was the most unpleasant moment of the decision, in which an error, if not allowed, then constantly appears. However, there is an effective verification technique here, which I talked about in class. Directional derivative and gradient.
All the “ingredients” have been found and now it’s a matter of careful substitution with further simplifications: 
– general equation the desired tangent plane.
I strongly recommend checking this stage of the solution as well. First you need to make sure that the coordinates of the tangent point really satisfy the found equation: 
- true equality.
Now we “remove” the coefficients general equation planes and check them for coincidence or proportionality with the corresponding values. In this case they are proportional. As you remember from analytical geometry course, - This normal vector tangent plane, and he is also guide vector normal straight line. Let's compose canonical equations normals by point and direction vector: 
In principle, the denominators can be reduced by two, but there is no particular need for this
Answer:
It is not forbidden to designate the equations with some letters, but, again, why? Here it’s already extremely clear what’s what.
The following two examples are for independent decision. A little “mathematical tongue twister”:
Example 2
Find the equations of the tangent plane and the normal to the surface at the point.
And a task that is interesting from a technical point of view:
Example 3
Write equations for the tangent plane and normal to the surface at a point
At the point.
There is every chance of not only getting confused, but also encountering difficulties when recording canonical equations of the line. And the normal equations, as you probably understand, are usually written in this form. Although, due to forgetfulness or ignorance of some nuances, the parametric form is more than acceptable.
Sample samples finalization of solutions at the end of the lesson.
Is there a tangent plane at any point on the surface? In general, of course not. The classic example is conical surface 
and point - the tangents at this point directly form a conical surface, and, of course, do not lie in the same plane. It is easy to verify that something is wrong analytically: .
Another source of problems is the fact non-existence any partial derivative at a point. However, this does not mean that at a given point there is no single tangent plane.
But it was, rather, popular science rather than practically significant information, and we return to pressing matters:
How to write equations for the tangent plane and normal at a point,
if the surface is specified by an explicit function?
Let's rewrite it implicitly:
And using the same principles we find partial derivatives: 
Thus, the tangent plane formula is transformed into the following equation:
And accordingly, the canonical normal equations:
As you might guess, 
- these are already “real” partial derivatives of a function of two variables at the point, which we used to denote by the letter “z” and were found 100500 times.
Please note that in this article it is enough to remember the very first formula, from which, if necessary, it is easy to derive everything else (of course, having basic level preparation). This is exactly the approach that should be used when studying the exact sciences, i.e. from a minimum of information we must strive to “draw” a maximum of conclusions and consequences. “Consideration” and existing knowledge will help! This principle is also useful because it will most likely save you in a critical situation when you know very little.
Let’s work out the “modified” formulas with a couple of examples:
Example 4
Write equations for the tangent plane and normal to the surface 
at point .
There is a slight overlay here with the notations - now the letter denotes a point on the plane, but what can you do - such a popular letter...
Solution: let’s compose the equation of the desired tangent plane using the formula:
Let's calculate the value of the function at the point:
Let's calculate 1st order partial derivatives at this point: 
Thus:
carefully, don't rush: 
Let us write down the canonical equations of the normal at the point: 
Answer:
And a final example for your own solution:
Example 5
Write down equations for the tangent plane and the normal to the surface at the point.
Final - because I have explained virtually all the technical points and there is nothing special to add. Even the functions themselves proposed in this task are dull and monotonous - in practice you are almost guaranteed to come across a “polynomial”, and in this sense, Example No. 2 with an exponent looks like a “black sheep”. By the way, it is much more likely to encounter a surface defined by an equation, and this is another reason why the function was included in the article as number two.
And finally, the promised secret: so how to avoid cramming definitions? (I, of course, do not mean the situation when a student is feverishly cramming something before an exam)
The definition of any concept/phenomenon/object, first of all, gives an answer to the following question: WHAT IS IT? (who/such/such/are). Consciously answering this question, you should try to reflect significant signs, definitely identifying a particular concept/phenomenon/object. Yes, at first it turns out to be somewhat tongue-tied, inaccurate and redundant (the teacher will correct you =)), but over time, quite decent scientific speech develops.
Practice on the most abstract objects, for example, answer the question: who is Cheburashka? It's not that simple ;-) This is " fairy tale character with big ears, eyes and brown fur"? Far and very far from definition - you never know there are characters with such characteristics... But this is much closer to the definition: “Cheburashka is a character invented by the writer Eduard Uspensky in 1966, who ... (listing of the main distinguishing features)”. Notice how well it started